By Nami Youn




 


Write-up  #8

Altitudes and Orthocenters





1. Construction

Given  a triangle ABC, I will construct the Orthocenter H of triangle ABC and then the Orthocenters of each triangle HBC, HAB, and HAC. The four circles are the circumcircles of triangelks ABC, HBC, HAB, and HAC.
 
 







Click here for a GSP sketch.


2. Examining the details

For an acute triangle ABC, let the point H be the orthocenter and points D, E, and F be the feet of the perpendiculars from A, B, and C, respectively.

Prove that

(1)  HD/AD + HE/BE + HF/CF = 1
(2)    AH/AD + BH/BE + CH/CF = 2
 
 








Proof.
 

(1)  HD/AD + HE/BE + HF/CF = 1

Let the area of ABC = S, are of HAB = S1, area of HBC = S2, and area of HCA = S3.  Then

S = 1/2 AB*CF = 1/2 BC*AD = 1/2 CA*BE

S1 = 1/2 AB*HF
S2 = 1/2 BC*HD
S3= 1/2 CA*HE

Since  S = S1 + S2 + S3, we can obtain

1 = (S1 + S2 + S3)/S = S1/S + S2/S + S3/S = HF/CF + HD/AD + HE/BE

Therefore,  HD/AD + HE/BE + HF/CF = 1
 
 
 

(2)    AH/AD + BH/BE + CH/CF = 2

Since HD = AD - AH, HE = BE - BH, and HF = CF - CH,  and by the fact (1)

    1  =  HD/AD + HE/BE + HF/CF
        =  (AD - AH)/AD + (BE - BH)/BE + (CF - CH)/CF
        = 3 - (AH/AD + BH/BE + CH/CF)
 

Therefore, AH/AD + BH/BE + CH/CF = 2


3. What if ABC is an obtuse triangle?

Let's consider an obtuse triangle. Do the facts (1) and (2) still hold?

In an obtuse triangle, the orthocenter is outside of the given triangle.  In the below figure, the point H is the orthocenter and therefore the relationship (1) and (2) cannot be true any more.
 
 







However, the point A is the orthocenter of triangle HBC.  So we can adjust the previous relationships as below.

(1)  AD/HD + AE/BE + AF/CF = 1
(2)  AH/HD + BA/BE + CA/CF = 2
 

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